Question

Calculate the standard enthalpy change for the precipitation reaction of 50.0 grams of barium chloride and 75.0 grams of potassium sulfate to form solid barium sulfate and aqueous potassium chloride, given that the heat of formation of solid barium sulfate is -1474 kJ/mol and aqueous potassium chloride is -436 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the precipitation reaction, we first need to determine the balanced chemical equation for the reaction:BaCl2 aq  + K2SO4 aq   BaSO4 s  + 2 KCl aq Next, we need to determine the moles of each reactant:Molar mass of BaCl2 = 137.3  Ba  + 2 * 35.45  Cl  = 208.2 g/molMoles of BaCl2 = 50.0 g / 208.2 g/mol = 0.240 molMolar mass of K2SO4 = 2 * 39.1  K  + 32.1  S  + 4 * 16.0  O  = 174.3 g/molMoles of K2SO4 = 75.0 g / 174.3 g/mol = 0.430 molNow, we need to determine the limiting reactant. The stoichiometry of the reaction is 1:1 for BaCl2 and K2SO4, so we can compare the moles of each reactant:0.240 mol BaCl2 < 0.430 mol K2SO4BaCl2 is the limiting reactant, and the reaction will produce 0.240 mol of BaSO4 and 0.480 mol of KCl.Now we can calculate the standard enthalpy change for the reaction using the heat of formation values:H = [moles of products * heat of formation of products] - [moles of reactants * heat of formation of reactants]Since the heat of formation for the reactants  BaCl2 and K2SO4  are not given, we will assume that they are in their standard states and their enthalpy of formation is zero. Therefore, the enthalpy change for the reaction will only depend on the products:H = [ 0.240 mol * -1474 kJ/mol  +  0.480 mol * -436 kJ/mol ] - [0]H =  -352.8 kJ  +  -209.3 kJ  = -562.1 kJThe standard enthalpy change for the precipitation reaction is -562.1 kJ.